# Orbits of unipotent groups

Let $$V$$ be a real vector space (of finite dimension) and let $$G$$ be a unipotent Lie subgroup of $$\mathrm{GL}(V)$$. The orbits of points under the action of $$G$$ (that is, the sets $$Gx = \{T(x) \ : \ T \in G\}$$ where $$x \in V$$) are algebraic varieties --- this follows from Kostant-Rosenlicht theorem. I am wondering whether more is true:

Question: Let $$x \in V$$. Do there exist polynomial maps $$f_1,\dots,f_s \colon V \to \mathbb{R}$$ whose zero locus is $$Gx$$ and which are invariant under the action of $$G$$? (i.e.: $$Gx = \{ y \in V \ : \ f_i(y) = 0 \text{ for } i = 1,2,\dots,s\}$$ and $$f_i(Ty) = f(y)$$ for all $$T \in G,\ y \in V$$.)

• No. Let the additive group act on the affine plane by $t\cdot (x,y)=(x,y+tx)$. Thus the orbits are the vertical lines $D_x=\{(x,t):t\in\mathbf{R}\}$ for $x\neq 0$, and all the singletons $\{(0,t)\}$. Hence the invariant polynomials are those that constant on vertical lines, i.e., those $P(x,y)=Q(x)$. Zero locus of a family of such polynomials is a union of vertical lines, and hence cannot be one of these invariant singletons. – YCor Jan 24 '19 at 19:28
• Oh, right. It was actually looking at this very action that led me to the hope that the orbits have to take the form as described above - only I forgot about the singletons. Still the vertical lines do have the desired form. Is that a coincidence, or does some variant of the above question have a positive answer (e.g. with additional assumptions on $x$ or on $G$, or both)? – Jakub Konieczny Jan 25 '19 at 15:00
• One can wonder whether there is an open $G$-invariant Zariski-dense $U$ subset, definable in terms of $G$-invariant polynomials, such that the $G$-invariant polynomials on $V$ separate $G$-orbits within $U$. I don't know if it's true: one should check examples such as those unipotent actions for which the algebra of invariant polynomials is not finitely generated. – YCor Jan 25 '19 at 15:57